Ch. 20 Thermal systems spontaneously ... - University of MiamiThermal systems spontaneously change...
Transcript of Ch. 20 Thermal systems spontaneously ... - University of MiamiThermal systems spontaneously change...
Thermal systems spontaneously change only in certain ways:
Spontaneous heat flow always occurs from a hotter body to a colder body
Ch. 20
2nd Law of Thermodynamics
(even though to do the reverse does not violate the 1st Law of Thermodynamics)
Reversible vs Irreversible Processes
Reversible Equilibrium (quasi-equilib.)Irreversible Nonequilibrium
Heat Engine: a device that converts heat to work
Simple engines involve a cyclic process of aworking substance (usually an ideal gas)
WQU =⇒=Δ 0Net heat flow into working substance equalsnet work done by engine
Two thermal reservoirs [one hot (H), one cold (C)]:exchange heat with working substanceat constant T (QH and QC)
Heat Engine: a device that converts heat to work
H
C
H
C
H QQ
QW
−=+== 11ηEfficiency of engine:
20.39 A heat engine takes 0.350 mol of a diatomic idealgas around the cycle shown.
What is the efficiency?
HQW
=η
TRnTnCQU V Δ⎟⎠⎞
⎜⎝⎛=Δ==Δ
25
Compute the W, Q for each process:
1 2: isochoric W=0 (dV=0)1st Law
JKmolKJmol
31018.2)300)(/314.8)(2/5)(350.0(
×=
=
2 3: adiabatic Q=0
)0(786)108)(/314.8)(2/5)(350.0(
>−=−⋅=
Δ=−=Δ
WJKKmolJmol
TnCWU V
3 1: isobaric
JKKmolJmol
TnRVpW
559)192)(/314.8)(350.0(
−=−⋅=
Δ=Δ=
RRCCTnCQ Vpp )2/7(, =+=Δ=
JKKmolJmolQ
1956)192)(/314.8)(2/7)(350.0(
−=−⋅=
Net Work done:
JJJWW 2275597861332 =−=+ →→
Heat Input (QH):
JQ 321 1018.2 ×=→
HQW
=η
%4.10104.01018.2
2273
==×
=J
Jη
Engine Statement of the 2nd Law
It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work, with the system ending up in the same state in which it began.
Or equivalently…
Heat cannot be converted completely into work with no other change taking place.
Or equivalently…
There are no engines with 100% efficiency.
Internal Combustion Engines
Otto Cycle
fuel injected
Diesel Cyclefuel injected
allows forgreater r,efficiency
Brayton Cycle (jet engines)p
V
rp
p Taab
c dab: isobaric compressionbc: adiabatic compressioncd: isobaric expansionda: adiabatic expansion
What is the efficiency?
ab:
)(
)(
bap
abpC
TTnC
TTnCQ
−−=
−=
H
C
H QQ
QW
−== 1η
cd: )( cdpH TTnCQ −=
express Tb, Tc, Td in terms of Ta, r
cd
ba
TTTT
−−
−=1η
QC
QH
p
V
rp
p Taab
c d
γγγγad TpTrp −− = 11)(
γγ /)1( −=r
TT ad
dbb
bbb
nRTVrpnRTVp=
=)(
γ/1rTT a
b =
rTT d
b =
)/1/1()/11(1 1/2/)1(
/1
−− −−
−= γγγ
γ
rrTrT
a
a
cd
ba
TTTT
−−
−= 1η
γγ /)1(1 −−= r
p
QC
QH
V
rp
p Taab
c d
γγ /)1( −=r
TT bc
1/2 −= γrTT a
c
r0 2 4 6 8 10 12
e
0.0
0.1
0.2
0.3
0.4
0.5
0.6
γγ /)1(1 −−= r
QC
QH
V
rp
p Taab
c d
Refrigerators: heat engine in reverse
CH
CC
CH
CH
QQQ
WQ
K
WQQWQQ
−==
+=
−=−
Coefficient of performance
Real refrigerators
2nd Law of Thermodynamics
It’s impossible to convert heatcompletely to work (η=1), with no other change taking place.
It’s impossible for any process to haveas its sole result the transfer of heat from a colder body to a hotter body.
Engine statement:
Refrigerator statement:
Equivalence of Engine and Refrigerator Statements of 2nd Law
Carnot Cycle: maximum efficiency
H
C
H
C
TT
−=−= 11η
No engine can be more efficient than a Carnot engineoperating between the same two temperatures.
Equivalent statement of the 2nd Law
Proof: Suppose there is a hypothetical engine more efficient than Carnot (designate with ′)
Take Carnot refrigerator with hypothetical super-efficient engine supplying the work and QC’s the same:
H
C
′−=′ 1η
H
C
−= 1η
HH
H
C
H
C
QQQQ
>′⇒
<′
⇒>′ ηη
Δ+=′ HH QQLet
CH QQW −′=′1st Law: CH QQ −Δ+= Δ+= W
Violation of 2nd LawNo engine can be more efficient than a Carnot engineoperating between the same two temperatures.
Entropy Statement of 2nd Law
In any thermodynamic process that proceeds from one equilibrium state to another, entropy of the system+environment (i.e. the “universe”) either remains unchanged or increases.
Equivalent to Engine statement: there are no perfect engines